Given:

The curve is y = 3x² – 9x + 8

Differentiating the above w.r.t x

⇒ y = 3x² – 9x + 8

⇒\(\frac{dy}{dx}\) = 2 x 3x^{2 – 1} – 9 + 0

⇒ \(\frac{dy}{dx}\) = 6x – 9 ...(1)

Since, the tangent are equally inclined with axes

i.e, θ = \(\frac{\pi}{4}\) or θ \(=\frac{-\pi}{4}\)

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

⇒ \(\frac{dy}{dx}\) = tan( \(\frac{\pi}{4}\) ) or tan( \(\frac{-\pi}{4}\) )

⇒ \(\frac{dy}{dx}\) = 1or – 1 ...(2)

\(\therefore\) tan( \(\frac{\pi}{4}\) ) = 1

From (1) & (2),we get,

⇒ 6x – 9 = 1 0r 6x – 9 = – 1

⇒ 6x = 10 0r 6x = 8

⇒ x = \(\frac{10}{6}\) or x = \(\frac{8}{6}\)

⇒ x = \(\frac{5}{3}\) or x = \(\frac{4}{3}\)

Substituting x = \(\frac{5}{3}\) or x = \(\frac{4}{3}\) in y = 3x² – 9x + 8,we get,

When x = \(\frac{5}{3}\)

⇒ y = 3( \(\frac{5}{3}\) )² – 9( \(\frac{5}{3}\) ) + 8

⇒ y = 3( \(\frac{25}{9}\) ) – ( \(\frac{45}{3}\) ) + 8

⇒ y = ( \(\frac{75}{9}\) ) – ( \(\frac{45}{3}\) ) + 8

taking LCM = 9

⇒ y = ( \(\frac{(75\times1)-(45\times3)+(8\times9)}{9}\) )

⇒ y = ( \(\frac{75-135+72}{9}\) )

⇒ y = ( \(\frac{12}{9}\) )

⇒ y = ( \(\frac{4}{3}\) )

⇒ y = 3( \(\frac{4}{3}\) )2 – 9( \(\frac{4}{3}\) ) + 8

⇒ y = 3( \(\frac{16}{9}\) ) – ( \(\frac{26}{3}\) ) + 8

⇒ y = ( \(\frac{48}{9}\) ) – ( \(\frac{36}{3}\) ) + 8

taking LCM = 9

⇒ y = ( \(\frac{(48\times1)-(36\times3)+(8\times9))}{9}\) )

⇒ y = ( \(\frac{48-108+72}{9}\) )

⇒ y = ( \(\frac{12}{9}\) )

⇒ y = ( \(\frac{4}{3}\) )

Thus, the required point is ( \(\frac{5}{3}\), \(\frac{4}{3}\)) & ( \(\frac{4}{3}\), \(\frac{4}{3}\))